TITLE: Simplifying $\sqrt{\underbrace{11\dots1}_{2n\ 1's}-\underbrace{22\dots2}_{n\ 2's}}$
QUESTION [14 upvotes]: How do I simplify:
$$\sqrt{\underbrace{11\dots1}_{2n\ 1's}-\underbrace{22\dots2}_{n\ 2's}}$$
Should I use modulos or should I factor them?  Or any I suppose to use combinatorics?  Any one have a clue?

REPLY [2 votes]: Observe that for integer $m\ge 0,$
   $$\underbrace{11\dots1}_{m\ 1's}=\frac{\underbrace{99\dots99}_{m\ 9's}}9=\frac{10^m-1}9 $$
$$\text{So, }\underbrace{11\dots11}_{2n\ 1's}=\frac{10^{2n}-1}9\text{ and }\underbrace{22\dots22}_{n\ 2's}=2\cdot \underbrace{11\dots11}_{n\ 1's}=2 \cdot\frac{10^n-1}9$$
$$\implies \underbrace{11\dots1}_{2n\ 1's}-\underbrace{22\dots2}_{n\ 2's}=\frac{10^{2n}-1-2(10^n-1)}9=\frac{10^{2n}-2\cdot10^n+1}9=\left(\frac{10^n-1}3\right)^2$$
$$\text{Now as }\underbrace{11\dots1}_{m\ 1's}=\frac{10^m-1}9,$$
$$ \frac{10^n-1}3=3\cdot \frac{10^n-1}9=3\cdot(\underbrace{11\dots11}_{n\ 1's})=\underbrace{33\dots33}_{n\ 3's}$$